Difference between revisions of "2024 AIME II Problems/Problem 7"
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| − | Let | + | ==Problem== |
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| + | Let <math>N</math> be the greatest four-digit positive integer with the property that whenever one of its digits is changed to <math>1</math>, the resulting number is divisible by <math>7</math>. Let <math>Q</math> and <math>R</math> be the quotient and remainder, respectively, when <math>N</math> is divided by <math>1000</math>. Find <math>Q+R</math>. | ||
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| + | ==Solution 1== | ||
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| + | We note that by changing a digit to <math>1</math> for the number <math>\overline{abcd}</math>, we are subtracting the number by either <math>1000(a-1)</math>, <math>100(b-1)</math>, <math>10(c-1)</math>, or <math>d-1</math>. Thus, <math>1000a + 100b + 10c + d \equiv 1000(a-1) \equiv 100(b-1) \equiv 10(c-1) \equiv d-1 \pmod{7}</math>. We can casework on <math>a</math> backwards, finding the maximum value. | ||
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| + | (Note that computing <math>1000 \equiv 6 \pmod{7}, 100 \equiv 2 \pmod{7}, 10 \equiv 3 \pmod{7}</math> greatly simplifies computation). | ||
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| + | Applying casework on <math>a</math>, we can eventually obtain a working value of <math>\overline{abcd} = 5694 \implies \boxed{699}</math>. ~akliu | ||
Revision as of 19:10, 8 February 2024
Problem
Let 
be the greatest four-digit positive integer with the property that whenever one of its digits is changed to 
, the resulting number is divisible by 
. Let 
and 
be the quotient and remainder, respectively, when is divided by 
. Find 
.
Solution 1
We note that by changing a digit to for the number 
, we are subtracting the number by either 
, 
, 
, or 
. Thus, 
. We can casework on 
backwards, finding the maximum value.
(Note that computing 
greatly simplifies computation).
Applying casework on , we can eventually obtain a working value of 
. ~akliu
